A certain circle can be represented by the following equation. $x^2+y^2+14x-10y+65=0$ What is the center of this circle ? $($
Answer: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+14x-10y+65&=0\\\\ x^2+y^2+14x-10y&=-65\\\\ (x^2+14x)+(y^2-10y)&=-65 \text{(rearrange terms)}\\\\ (x^2+14x{+49})+(y^2-10y{+25})&=-65{+49}{+25}\end{aligned}$ Notice that we must add ${49}$ and ${25}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 49 and 25?] Writing the equation in standard form $\begin{aligned}(x^2+14x{+49})+(y^2-10y{+25})&=-65{+49}{+25}\\\\ (x+7)^2+(y-5)^2&=9\\\\ (x-(-7))^2+(y-5)^2&=3^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(-7,5)$ and has a radius of $3$ units. Summary The circle is centered at $(-7,5)$. The circle has a radius of $3$ units.